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JEE Main 4 Apr 2024 Shift 2 Solved Paper
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© examsnet.com
Question : 15
Total: 90
If the mean of the following probability distribution of a random variable
X
;
X
0
2
4
6
8
P(X)
a
2a
a+b
2b
3b
is
46
9
, then the variance of the distribution is
[4 Apr 2024 Shift 2]
581
81
566
81
173
27
151
27
Validate
Solution:
👈: Video Solution
∑
P
i
=
1
a
+
2
a
+
a
+
b
+
2
b
+
3
b
=
1
4
a
+
6
b
=
1
. . . (I)
E
(
x
)
=
mean
=
46
9
∑
P
i
X
i
=
46
9
⇒
4
a
+
4
a
+
4
b
+
12
b
+
24
b
=
46
9
8
a
+
40
b
=
46
9
4
a
+
20
b
=
23
9
.
.
.
.
(II)
Subtract (I) from (II) we get
b
=
1
9
&
a
=
1
12
Variance
=
E
(
x
i
2
)
−
E
(
x
i
)
2
E
(
x
i
2
)
=
0
2
×
9
2
+
2
2
×
2
a
+
4
2
(
a
+
b
)
+
6
2
(
2
b
)
+
8
2
(
3
b
)
=
24
a
+
280
b
Put
a
=
1
12
b
=
1
9
E
(
x
i
2
)
=
2
+
280
9
=
298
9
∴
σ
2
=
E
(
x
i
2
)
−
E
(
x
i
)
2
=
298
9
−
(
46
9
)
2
σ
2
=
298
9
−
2116
81
=
566
81
© examsnet.com
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